# Full text of "A Novel Look at the One Dimensional Dirac Delta Schrodinger Equation"

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```A Novel Look at the One Dimensional
Delta Schrodinger Equation

By

Patrick Bruskiewich

Department of Physics and Astronomy,

University of British Columbia

email: patrickb@physics . ubc . ca

Resume

In this paper both integral transforms and an algebraic reformulation
of the second order differential equation are used to derive the solution to
the one dimensional attractive delta Schrodinger equation. This is a novel
the reformulation methods provide additional insight into the nature of
the one dimensional attractive delta potential.

Introduction

In many problems in quantum mechanics one can arrive at a solution in
several ways. It is sometimes interesting to revisit a familar problem to
see what new can be learned. Both algebraic and analytic techniques are
used in this paper to solve for the one dimensional Schrodinger equation
with an attractive delta potential.

An attractive delta potential can be used to model a point particle, one
dimensional models of molecules with multiple centres and in the Kronig-
Penney model of crystalline structure to name just a few occurences of the
potential in the literature. It is also of central importance in the Green's
Function formulation of quantum mechanics.

22

Gamma 138 Patrick Bruskiewich

The traditional manner in which to work with the one dimensional
Schrodinger equation is through differentiation. The fundamental equa-
tions of quantum mechanics can also be derived from a variational prin-
ciple or by means of integration. This has direct bearing on the manner
in which the solution to the one dimensional attractive delta Schrodinger
equation is derived in this paper.

We must begin by being mindful that the delta function is not a func-
tion in the traditional sense. It is a generalized function, or distribution,
which can be thought of as a limit of a sequence of functions of ever
increasing height and ever-decreasing width, used as an integrand in an
integral equation.

For example in its simplest representation, we can model an attractive
delta potential by a square well potential in the limit that its depth (Vo)
approaches infinity while its width (a) approaches but never reaches zero
in the limit, (with the proviso that Vo a ~ 1 in the limit).

In the limit we can also represent a delta function in several other
relevant ways, namely

s{x) = 7?^7? exp[ -7

5(x) = — lim

7T x->0 x 2 + e 2

_, . 1 siniVa:
dlx) = — lim

As is evident, each of these representations has a characteristic width
which approaches, but never reaches zero in their limit. This is of impor-
tance when we remember that a physical state must always correspond
to a square-integrable wavefunction.

If 6(x — a) represents a delta function at point a then given some
function f(x)

23

A Novel Look at the One Dimensional Delta Schrodinger Equation Gamma 138

/OO rOO

f(x)5(x — a)dx = fia) / 6(x — a)dx = fid)
-00 ' J — OO

The standard interpretation of the delta function, irrespective of its
functional representation, is that under the integral sign the delta func-
tion serves to "pick out" the value of the function fix) at the point a.

Consider the one dimensional Schrodinger equation with an attractive
delta potential given by

H 2 d 2

ip(x) — a 6(x) ip(x) = E ift(x).

2m dx 2

The standard way to solve this equation is to match continuity condi-
tions. For a potential centred on x = that spans — oo < x < +oo we
consider both the Dirichlet boundary condition and the Neumann bound-
ary condition. [1]

Let us consider the representation of the delta function as a box of
small width 2e centred on x = and large negative potential —a. In the
region x < — e, the potential is zero so we have a solution of the form

ift-(x) = Aex.p(nx)

where k = ^ — ^ .

Similarly in the region x > e the potential is again zero and in this
region we have a solution of the form

ip+(x) = .Bexp(— kx)
24

Gamma 138 Patrick Bruskiewich

It remains only to match these two functions at x = with the bound-

dx

ary conditions, namely ip is continuous and -f- is continuous except at

points where the potential is infinite.

Evidently the delta function must describe the discontinuity in the
derivative of ip at x = 0. We can extract the delta function by integrating
the Schrodinger equation, from — e to +e, and then take the limit as e — > 0,
namely

/ !T~2 ^( x ) dx — a 5(x) ip(x) dx = E ^{x) dx.

In the limit the last integral is clearly equal to zero, while the first
integral is

/: S «•> - - *2>

which is a measure as to how much the slope in the wavefunction changes
over the interval — e < x < e.

Thus we have simplified the expression to

.,dib. —2ma .. r+e —2ma , , n .

A( i ) = -&- H L *« *w dx = n^ W)

Through integration we have reduced the order of the Schrodinger
equation from second to first order and we incorporated the boundary
conditions in the first order equation explicitly.

By symmetry we recognize that

Solving this first order differential equation we arrive at the familiar
normalized wavefunction solution given by

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A Novel Look at the One Dimensional Delta Schrodinger Equation Gamma 138

, . y/2ma . ma I x I

* (x) = -jr~ exp( — w~ ]

The energy of this normalized wavefunction solution is given by

ma 2

£ = ~w

There are several other ways in which to derive the solution to the one
dimensional Schrodinger equation with an attractive delta potential. The
alternate techniques outlined in this paper can act as a starting point to
describing the quantum behaviour of other systems.

Solution by an Integral Transform

Consider once again the one dimensional attractive delta Schrodinger
equation

h 2 d 2

ip — a 5(x) tjj = E tp.

2m dx 2

We can borrow a technique from quantum field theory and derive a so-
lution using an integral transform in the way of a Fourier transform. [2]
Let

1 °r

^(x) = — dk e lkx cf>(k)

— 00

where <j){k) is the wave function in the momentum representation. Intro-
ducing this solution into the differential equation we find that

(^ k 2 + B) m = a V>(0)
26

Gamma 138

Patrick Bruskiewich

where "0(0) is the wavefunction value at x = and B
of the eigensolution. Solving for <j){k) we find

E, is the energy

m

a

*(0)

B 2 + B

Using the completeness theorem,

00 00 00

1 = I \$*{x) if)(x) dx = J J

dk dk'

J(k-k/)x ±*( h i

— 00

— 00 —00

2tt 2tt

WW*)

which when integrated over k' simplifies to

-i 00

— / dk (f)*(k) <f>(k)

— 00

2maip(ti)

h 2

-i 00

- /-

7T i

dk

[fc 2 +

2mB ] 2

where we recognize that the integrand is an even function and change
the integral limits accordingly. Upon integration we arrive at an inverse
tangent function, namely

tf

2maijj(0)_

2 7T

n i 3 r kh - k=0 °

arctan

V2mB

V2mB

k=0

which when we solve for E yields (E = — B)

E = -

ma

2h 2

We can use our value for B in the wavefunction (j){k) and solve for ip(x)
in the transform equation, namely

i/)(x) =

2ma V>(0)

h 2

°r dk

-Akx

-oo

2tt k 2 +[

ti 2

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A Novel Look at the One Dimensional Delta Schrodinger Equation Gamma 138

By inspection we see that only the even function cos(A;a:) yields a non-
zero contribution, and so adjusting the limits of integration accordingly
we find

*(*) = — si— J-

2ma "0(0) °f dk cos(kx)

k 2 + [f^ 12

which yields the solution

il){x) = V>(0)exp( — — )

Solution by Reformulation

We can find a solution to the one dimensional attractive delta Schrodinger
equation by using a simple transformation to the second order differential
equation. [3]

Consider the delta function S(x) represented by the second partial
derivative of the absolute value function in x, namely

1 d 2
*(*) = 2W I X I •

Gathering the terms in the second derivative we can rewrite the Schrodinger
equation as

h " d " ! a r Q2 i 11/ w ,

2m dx 2 2 dx 2

Bringing the wavefunction ip over to the left hand side, we get

1 d ,dib s ma d ,d\x\ s 2m I E

ip dx dx h 2 dx dx h 2

28

Gamma 138 Patrick Bruskiewich

We have used the fact that for a bounded wavefunction the energy E
is less than zero, namely

E = -\E\.

Bringing the wavefunction ip into the bracketed derivative and gather-
ing terms we have

d , 1 dip ma d I x L 1 , dip . 9 2ra I E I

— ( — H ! — -) H (— -) = ! -■

dx ip dx H 2 dx ip 2 dx h 2

Let us set the following condition on tp

d A dip ma d \ x \
dx ip dx h 2 dx
Solving this equation we get

^ = ■0(0) exp( — — | x |).

Using the normalization condition we find the familiar wavefunction
for the delta potential

. J ma , ma . ..

^ = __ exp( __ | x p.

Having found ip we can now use this wavefunction to solve for the
remaining condition, namely the value for the energy of the wavefunction,

1 M.n 2m I E
"(-ST) =

ip 2 dx h 2

Solving for the Energy (| E |= — E we get

„ ma 2

E =

2h 2 '

which is the expected value for the energy of a bounded wavefunction for
the delta potential.

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A Novel Look at the One Dimensional Delta Schrodinger Equation

Gamma 138

Solution by a Laplace Transform

Consider once again the one dimensional attractive delta Schrodinger
equation

2 £>2

K l d

2m dx 2

ip — a 5(x) ip = E i\).

We can solve for this second order differential equation using the Laplace
transform. Applying the usual transform

00

L(f) = Jf(t)e- st dt

to this second order differential equation yields

( s 2 + -ft- ) L (VO + ( -^- - s) V^(O) - <//(0) =
Let i/j(0) = C and ^'(0) = 7 so that

9 2mE 2ma

" ' ) L (VO + ( -r^- - s ) C - 7

(* 2 +

ft 2

ft 2

Solving for L (*0) provides the following transform equation which
needs to be solved:

MVO

(7-^0

( s 2 + b 2 ) ' ( s 2 + b 2 )

*c

+

where 6 2 = -^-. The inverse transform of the first term is

L- 1

(s 2 + b 2 )

( cos ibx)

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Gamma 138

Patrick Bruskiewich

If we let r\ = (7 — ^p C) then the inverse transform of the second
term is of the form

r\ L

-1

2ib (s-ib) 2ib (s + ib)

V

2ib

kl

Jbx

—ibx

]

1

b

sin

(bx)

So the solution for a one dimensional attractive delta Schrodinger equa-
tion using the Laplace transform method is

tjj(x)

Tj

C cos (bx) + — sin (bx)

( cos (bx) +

1

7

2ma

c

sin (bx)

which appears of a very different form to that of the solution given in part
one of this paper. However, it is possible to rewrite this eigensolution in
a more familiar form.

If we have a bounded solution we can change E — >

E I giving

ip(x) = £ cosh (| b I x) —
= C cosh (| b I x) +

77

sinh (| b I x)

7

2ma

c

sinh (| b I x)

since b -> i \ b |, cos (bx) — > cosh (| b \ x) and sin (bx) — >• i sinh (| 6 | #)
under this tranformation.

Since we are in the domain < x < 00 in our solution we need to
reformulate that part of cosh (bx) to conform to our boundary conditions
as # — > 00. We can do this by recognizing that

C exp (— I b I x) = ( [ cosh (| b \ x) — sinh (| b \ x)

m \a\

At this point, analysis of the sinh (bx) term shows that \ b \ n ,
that this term can be rewritten as

so

31

A Novel Look at the One Dimensional Delta Schrodinger Equation

Gamma 138

1

1 b\

7 -

2ma

h 2 C J

7 -

ma

h 2 C J

sinh (|

sinh (| b | x) =

sinh (| b | x) + (, sinh (| b \ x)

We see then we have

tp(x) = £ exp (— | b | x) +

ma

c

7

sinh (| b | x)

where the first term is the familiar solution to the one dimensional at-
tractive delta Schrodinger equation. There is still the second term, the
sinh (| b | x) part of the solution, we need to interpret.

Recalling that ift'(0) = 7 we see that the term in sinh(| b \ x) is, in
fact

1

T

ma

C - 7

sinh (| b I x) =

c -

<//(0)

sinh (| b I x)

By inspection we know that as x — > 00, the function sinh(| b \ x) — >
00. With some thought, however, and a careful consideration of the sign,
in light of the | b \ term in the expression

c

V/(0)

one sees immediately that we recover the well known solution to the
one dimensional Schrodinger equation with an attractive delta potential,
namely

, . y/2ma , ma I x L
tp{x) = — r — exp( — )

h

h 2

32

Gamma 138 Patrick Bruskiewich

Discussion

in the literature, as is the Laplace Transform approach. The use of the
Laplace transform to solve one dimensional equations in quantum me-
chanics is an under-utilized technique which should be explored fur-
ther. The direct mathematical tie between the Laplace transform and
the Fourier transform method is not so obvious, but is rich in implication.

Along with having applicability in quantum mechanics, the one dimen-
sional attractive delta potential is a good starting point to the study of
the Dirac delta potential in more than one dimension. This multidimen-
sional potential in turn allows the introduction within the framework of
quantum mechanics of many of the basic concepts of quantum field the-
ory, such as regularization, the renormalization group and dimensional
transformation.

References

[1] The standard derivation of the solution to the one dimensional attrac-
tive delta Schrodinger equation can be found in most any quantum
textbook, such as the texts by Dirac, Messiah, Merzbacher, Grif-
fiths or Gasiorowicz.

[2] For a succinct overview of this technique refer to section 10.8 of the
book Quarks, Leptons and Gauge Fields by Kerson Huang, World
Scientific, Singapore, 1982.

[3] The idea for this approach derives from a comment in section 6.2
of Quantum Mechanics, 2nd Edition by Eugen Merzbacher, John
Wiley and Sons, New York, 1970.

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